After all, both these statements do the same thing...
int a = 10;
int *b = &a;
printf("%p\n",b);
printf("%08X\n",b);
For example (with different addresses):
0012FEE0
0012FEE0
It is trivial to format the pointer as desired with %x, so is there some good use of the %p option?
x is Unsigned hexadecimal integer ( 32 Bit )
p is Pointer address
See printf on the C++ Reference. Even if both of them would write the same, I would use %p to print a pointer.
You cannot depend on %p displaying a 0x prefix. On Visual C++, it does not. Use %#p to be portable.
x is used to print t pointer argument in hexadecimal.
A typical address when printed using %x would look like bfffc6e4 and the sane address printed using %p would be 0xbfffc6e4
At least on one system that is not very uncommon, they do not print the same:
~/src> uname -m
i686
~/src> gcc -v
Using built-in specs.
Target: i686-pc-linux-gnu
[some output snipped]
gcc version 4.1.2 (Gentoo 4.1.2)
~/src> gcc -o printfptr printfptr.c
~/src> ./printfptr
0xbf8ce99c
bf8ce99c
Notice how the pointer version adds a 0x prefix, for instance. Always use %p since it knows about the size of pointers, and how to best represent them as text.
When you need to debug, use printf with %p option is really helpful. You see 0x0 when you have a NULL value.
The size of the pointer may be something different than that of int. Also an implementation could produce better than simple hex value representation of the address when you use %p.
Source: Stackoverflow.com